Given a univariate AR1 model with step size $n$

$$X(t+n) = a_n + b_n X(t) + \sigma_n Z(t+n),$$

we can change to a model with a different step size $m$

$$X(t+m) = a_m + b_m X(t) + \sigma_m Z(t+m),$$


$$ b_m = b_n^{m/n} $$ $$ a_m = a_n \frac{1 - b_m}{1 - b_n} $$ $$ \sigma_m = \sigma_n \sqrt\frac{1 - b_m^2}{1 - b_n^2} $$

But what if we have a multivariate model with, say, two AR1 processes $X(t)$ and $Y(t)$ where the innovations have a correlation of

$$Corr[Z_X(t+n),Z_Y(t+n)] = \rho_n$$

Is this correlation parameter affected by the step size change? How can we get from $\rho_n$ to $\rho_m$?

Add 1

The relationship between parameters for the univariate model are obtained by equating expressions for the unconditional mean, variance, and auto-correlation, i.e.

$$ E[X(t)] = \frac{a_n}{1 - b_n} = \frac{a_m}{1 - b_m} $$ $$ Var[X(t)] = \frac{\sigma_n^2}{1 - b_n^2} = \frac{\sigma_m^2}{1 - b_m^2} $$ $$ Cov[X(t), X(t - m \times n)] = b_n^m Var[X(t)] = b_m^n Var[X(t)] $$

Add 2

I suppose, similarly to the uni-variate derivation above, one can equate the expressions for cross-covariance between the two processes, i.e.

$$ Cov[X(t),Y(t)] = \frac{\rho_n \sigma_{x;n} \sigma_{y;n} }{1 - b_{x;n} b_{y;n}} = \frac{\rho_m \sigma_{x;m} \sigma_{y;m} }{1 - b_{x;m} b_{y;m}} $$

which would imply that

$$ \rho_m = \rho_n \frac{\sigma_{x;n} \sigma_{y;n}}{\sigma_{x;m} \sigma_{y;m}} \frac{1 - b_{x;m} b_{y;m}}{1 - b_{x;n} b_{y;n}} = \rho_n \frac{1 - b_{x;m} b_{y;m}}{1 - b_{x;n} b_{y;n}} \sqrt\frac{\left(1 - b_{x;n}^2\right)\left(1 - b_{y;n}^2\right)}{\left(1 - b_{x;m}^2\right)\left(1 - b_{y;m}^2\right)} $$

but I am not sure if this is the right way to do it.

  • $\begingroup$ Hi: could you explain or show the derivation of the $m$ parameters ? If you do that, someone might know how to do the correlation. Although someone might also know without that being shown :). $\endgroup$
    – mlofton
    Commented Aug 25, 2020 at 12:27
  • $\begingroup$ It's not clear to me how you obtained the expression for the cross correlation of X(t) and Y(t) but, assuming it's correct, I would think that that way would be the way to go. You seem to have answered your own question :). $\endgroup$
    – mlofton
    Commented Aug 26, 2020 at 12:30


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